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Math Help - Compex number question

  1. #1
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    Question Compex number question

    having trouble with this question:

    Given that p and q are real and that 1 + 2i is a root of this equation:

    z^2 + (p+5i)*z +q(2 - i) = 0

    - Determine the values of p and q
    - Determine the other root of the equation

    I really have no idea how to approach this.
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  2. #2
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    Quote Originally Posted by dojo View Post
    having trouble with this question:

    Given that p and q are real and that 1 + 2i is a root of this equation:

    z^2 + (p+5i)*z +q(2 - i) = 0

    - Determine the values of p and q
    - Determine the other root of the equation

    I really have no idea how to approach this.
    The other root will be the complex conjugate, you will need this to find p and q
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  3. #3
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    Quote Originally Posted by dojo View Post
    Given that p and q are real and that 1 + 2i is a root of this equation: z^2 + (p+5i)*z +q(2 - i) = 0
    - Determine the values of p and q
    - Determine the other root of the equation
    Quote Originally Posted by pickslides View Post
    The other root will be the complex conjugate, you will need this to find p and q
    No pickslides that is not true. It is true only if the coefficients are real.
    In this case they are not real.

    @ dojo
    Expand (1+2i)^2+(p+5i)(1+2i)+q(2-i) to get (p+q-13)+i(2p-2q+9).
    Now you have two equations in two unknowns.
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  4. #4
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    Hi plato - should that expansion not be (p + 2q - 13) + i(2p -q +9).

    I'm still at a loss - I thought it would be solved by comparing the original equation to (x - α) (x - β). After expanding with replacing with the root and its conjugate fianlly compare the coefficiants.

    I then get p = -2 -5i and q = 2 + i
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  5. #5
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    Quote Originally Posted by dojo View Post
    Hi plato - should that expansion not be (p + 2q - 13) + i(2p -q +9).

    I'm still at a loss - I thought it would be solved by comparing the original equation to (x - α) (x - β). After expanding with replacing with the root and its conjugate fianlly compare the coefficiants.

    I then get p = -2 -5i and q = 2 + i
    Your expansion is correct. But since you're told that p and q are real it is unlikely that your answers are correct ....

    Did you equate the real and imaginary parts of the expanded expression to zero and solve the resulting two equations simultaneously (as advised by Plato)? If so, please show your work.
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  6. #6
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    I'm lost Mr Fantastic. can you help me further as I dont know what you mean. Here are my workings...

    D
    Attached Thumbnails Attached Thumbnails Compex number question-complex-001.jpg  
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  7. #7
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    Quote Originally Posted by dojo View Post
    having trouble with this question:

    Given that p and q are real and that 1 + 2i is a root of this equation:

    z^2 + (p+5i)*z +q(2 - i) = 0

    - Determine the values of p and q
    - Determine the other root of the equation

    I really have no idea how to approach this.

    p + 2q - 13 + i(2p -q +9) = 0 + 0i.


    Equating real and imaginary parts gives:

    p + 2q - 13 = 0 and 2p - q + 9 = 0.


    Solve these equations simultaneously for p and q.
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  8. #8
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    Quote Originally Posted by dojo View Post
    I'm lost Mr Fantastic. can you help me further as I dont know what you mean. Here are my workings...

    D
    Your working is not correct

    Have another look at Plato's post

    Quote Originally Posted by Plato View Post
    No pickslides that is not true. It is true only if the coefficients are real.
    In this case they are not real.
    The two roots are not conjugate since the coefficients (p+5i and q(2-i)) are not real

    Quote Originally Posted by Plato View Post
    Expand (1+2i)^2+(p+5i)(1+2i)+q(2-i) to get (p+q-13)+i(2p-2q+9).
    Now you have two equations in two unknowns.
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  9. #9
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    OK... When I expand I get:

    (p + 2q + 15) + i(2p + 9 - q) = 0
    (p + 2q -13) + i(9 + 2p -q) = 0

    not sure how i would solve using simultaneous eq. though
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  10. #10
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    Expanding (1+2i)^2+(p+5i)(1+2i)+q(2-i)=0 gives (p+2q-13)+i(2p-q+9)=0

    Which gives two equations
    p+2q-13=0 (real part)
    2p-q+9=0 (imaginary part)
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  11. #11
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    Ah-ha I see. so I get p = -1 and q = 7.?

    What about finding the other root?
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  12. #12
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    You have to know that the roots of a quadratic az˛+bz+c=0 are linked by

    z_1 + z_2 = -\frac{b}{a}

    z_1 \times z_2 = \frac{c}{a}

    You can use either of these relations (the first one is the simplest one)
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